How do i get from one equation to another equation?

Question

((tan α+tan β)/(tan α-tan β)) to ((sin(α+β))/(sin(α-β)))

Answer 1

Tan α = Sin α / Cos α
Cot α = Cos α / Sin α

Tan α + Tan β = (Sinα/Cosα) + (Sinβ/Cosβ)

Now we have different denominators (Cosα and Cosβ)
Try to make a Greatest Common Divisor in the denominator:

Greatest Common Divisor between Cosα and Cosβ is Cosα * Cosβ so:


( (Sinα * Cosβ) + (Sinβ * Cosα) ) / (Cosα * Cosβ)

I think you can do the other equation in the denominator yourself.


Good luck.

Answer 2

We're getting sin(α+β) and sin(α-β) showing up in the end, so chances are we'll have to apply the addition and subtraction identities for sine at some point:
sin(a+b) = sin(a)cos(b) + sin(b)cos(a)
sin(a-b) = sin(a)cos(b) - sin(b)cos(a)

Start with the left and split up the tangents into sine/cosine:

(tan(α) + tan(β)) / (tan(α) - tan(β))
(sin(α)/cos(α) + sin(β)/cos(β)) / (sin(α)/cos(α) - sin(β)/cos(β))

Multiply the top and bottom by cos(α)cos(β)

Answer 3

Top line , N
sin α / cos α + sin β / cos β
(sin α cos β + sin β cos α ) / (cos α cos β)
sin (α + β) / cos α cos β

Bottom Line , D
sin (α - β) / (cos α cos β)
(sin α cos β - sin β cos α ) / (cos α cos β)
sin (α - β) / cos α cos β

N/D = sin (α + β) / sin (α - β)

Answer 4

let me assume alpha to be A and beta B....
tanA =sinA/cosA......... tanB=sinB/cosB
take lcm for tan A +tanB we get [sinAcosB+cosAsinB]/[cosA cosB]......(1) the numerator is precisely sin(A+B)

similarly tanA-tanB=[sinAcosB -cosAsinB]/cosAcosB.......(2) whose numerator is sin(A-B)

divide (1) by (2) to get the required answer...{ observe that cosAcosB gets cancelled when you divide....}

Answer 5

First convert tan(alpha) and tan(beta) into sin(al)/sin(be)
then you get
(sin(al)cos(be)+cos(al)sin(be))/(sin(al)cos(be)-cos(al)sin(be))
=sin(al+be)/sin(al-be) .....here al=alpha,be=beta